Question

If a quadrilateral ABCD is such that $\overrightarrow{AB}=b,\overrightarrow{AD}=d$ and $\overrightarrow{AC}=pb+qd(p+q\ge 1),$ then show that the area of the quadrilateral is $\frac{1}{2}(p+q)|b\times d|.$

Answer 1

Area $={\Delta }_1+{\Delta }_2$
$=\frac{1}{2}|b\times (pb+qd)|+\frac{1}{2}|(pb+qd)\times d|$
$=\frac{1}{2}\left|q(b\times d)\right|+\frac{1}{2}\left|p(b\times d)\right|$
$=\frac{1}{2}\left|(p+q)\right|\left|(b\times d)\right|$
$=\frac{1}{2}(p+q)|b\times d|$ ($\because p+q\ge 1$ so positive only).